Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $y = \dfrac{q + 10}{9q + 27} \times \dfrac{q^2 + q - 6}{-q - 10} $
Explanation: First factor the quadratic. $y = \dfrac{q + 10}{9q + 27} \times \dfrac{(q + 3)(q - 2)}{-q - 10} $ Then factor out any other terms. $y = \dfrac{q + 10}{9(q + 3)} \times \dfrac{(q + 3)(q - 2)}{-(q + 10)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (q + 10) \times (q + 3)(q - 2) } { 9(q + 3) \times -(q + 10) } $ $y = \dfrac{ (q + 10)(q + 3)(q - 2)}{ -9(q + 3)(q + 10)} $ Notice that $(q + 10)$ and $(q + 3)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ \cancel{(q + 10)}(q + 3)(q - 2)}{ -9\cancel{(q + 3)}(q + 10)} $ We are dividing by $q + 3$ , so $q + 3 \neq 0$ Therefore, $q \neq -3$ $y = \dfrac{ \cancel{(q + 10)}\cancel{(q + 3)}(q - 2)}{ -9\cancel{(q + 3)}\cancel{(q + 10)}} $ We are dividing by $q + 10$ , so $q + 10 \neq 0$ Therefore, $q \neq -10$ $y = \dfrac{q - 2}{-9} $ $y = \dfrac{-(q - 2)}{9} ; \space q \neq -3 ; \space q \neq -10 $